Problem: Simplify the following expression and state the condition under which the simplification is valid. $r = \dfrac{-6y^3 + 78y^2 - 216y}{5y^3 + 20y^2 - 160y}$
Explanation: First factor out the greatest common factors in the numerator and in the denominator. $ r = \dfrac {-6y(y^2 - 13y + 36)} {5y(y^2 + 4y - 32)} $ $ r = -\dfrac{6y}{5y} \cdot \dfrac{y^2 - 13y + 36}{y^2 + 4y - 32} $ Simplify: $ r = - \dfrac{6}{5} \cdot \dfrac{y^2 - 13y + 36}{y^2 + 4y - 32}$ Since we are dividing by $y$ , we must remember that $y \neq 0$ Next factor the numerator and denominator. $ r = - \dfrac{6}{5} \cdot \dfrac{(y - 4)(y - 9)}{(y - 4)(y + 8)}$ Assuming $y \neq 4$ , we can cancel the $y - 4$ $ r = - \dfrac{6}{5} \cdot \dfrac{y - 9}{y + 8}$ Therefore: $ r = \dfrac{ -6(y - 9)}{ 5(y + 8)}$, $y \neq 4$, $y \neq 0$